A magnetizing field of 500 A/m produces a magnetic flux of 4×10^(-5) Wb in an iron rod of cross-sectional area 0.4cm^2 . The permeability of the rod in W/b/A-m is

Correct answer: (C) 2×10^(-4). Given, magnetising field, H = 5000 A/m Magnetic flux, ϕ=4×10^(-5)Wb Cross-sectional area, A = 0.4 cm^2=4×10^(-5)m^2 ∴ Permeability of rod, μ=B/H=(ϕ)/AH∴ϕ=BA⇒B=(ϕ)/A =(4×10^(-5))/(4×10^(-5)×5000)=2×(10^(-4)Wb)/A-m

Q141 / 150 · MHT CET 2019 (Shift 1) · Physics

A magnetizing field of 500 A/m produces a magnetic flux of $4 \times 10^{- 5}$ Wb in an iron rod of cross-sectional area $0.4 c m^{2}$ . The permeability of the rod in W/b/A-m is

A
$4 \times 10^{- 6}$
B
$1 \times 10^{- 3}$
Correct answer
C
$2 \times 10^{- 4}$
D
$3 \times 10^{- 5}$

Solution

Given, magnetising field, H = 5000 A/m
Magnetic flux,

ϕ = 4 \times 10^{- 5} W b

Cross-sectional area, A = 0.4

c m^{2} = 4 \times 10^{- 5} m^{2}

∴

Permeability of rod,

μ = \frac{B}{H} = \frac{ϕ}{A H} [∴ ϕ = B A \Rightarrow B = \frac{ϕ}{A}]

= \frac{4 \times 10^{- 5}}{4 \times 10^{- 5} \times 5000} = 2 \times \frac{10^{- 4} W b}{A} - m

Hence, the correct answer is option: 'C'.

A magnetizing field of 500 A/m produces a magnetic flux of 4×10-5 Wb in an iron rod of cross-sectional area 0.4cm2 . The permeability of the rod in W/b/A-m is

Solution

More Magnetic Properties of Matter Questions

A magnetizing field of 500 A/m produces a magnetic flux of $4 \times 10^{- 5}$ Wb in an iron rod of cross-sectional area $0.4 c m^{2}$ . The permeability of the rod in W/b/A-m is